7.1 z=1.732 < 1.96 we accept the alternative hypothesis at the 5% level

    1. p-value = 0.083

    2. 0.1< p <0.2

    3. t=1.155

    4. power of the test = .46

    5. power of the test = 0.96

    6. n = 229

    7. (0.82,1.58)

    8. Since the interval contains 1.0 an4

    9. z=-2.005 < z(0.025) =-1.96 We reject the null hypothesis

    1. Null Hypothesis: p = p0

      Alternative Hypothesis: p not= p0

      Where p0 is the proportion of current smokers in the general population and p is the proportion of current smokers in the volunteer population.

    2. z=-4.36 and p value is <0.001. Therefore we consider the results to be highly significant.

    3. t(23,0.9995)= 3.767

    4. (160.2,189.8) Does not contain 230 so we can conlude that the underlying cholesterol level for macrobiotics is significantly lower than 230.

    5. The hypothesis test tells us precisely how significant the results are. The CI give us a range of values within which the true mean cholesterol for macrobiotics is likely to fall.

    1. Null Hypothesis: p = 0.12

      Alternative Hypothesis: p not= 0.12

    2. Two sided test.

    3. p = 0.165. There is no significant excess or deficit in the proportion of deaths due to lung cancer in this plant

    4. z = 2.66 and p = 0.008. There is a significant excess in the proportion of deaths due to lung cancer among workers in the four plants.

    5. z = 1.355 and p = 0.175. There is no significan excess or deficit onece deaths due to ischemic heart disease are excluded.

      7.80 z=-2.98 and p = 0.003. Rate of cancer in group A is significantly different from that of the general population.

      7.81 CI=(0.020,0.052)